COMEDK 2024 Shift 2 Paper

Let's consider the torque (also called the couple) acting on a bar magnet in a uniform magnetic field. The torque τ experienced by a bar magnet of magnetic moment M in a uniform magnetic field B, when held at an angle θ to the field, is given by:
τ=MBsin‌θ
Initially, the bar magnet is held perpendicular to the magnetic field, so θ=90∘. The torque in this position is:
τ0=MBsin‌90∘=MB
To halve the couple, the new torque τ should be:
τ=‌

τ0

2

=‌

MB

2

Let the new angle be θ. The torque at this angle will be:
τ=MBsin‌θ
Setting this equal to half the initial torque, we get:
MBsin‌θ=‌

MB

2

Dividing both sides by MB :
sin‌θ=‌

1

2

The angle θ for which sin‌θ=‌

1

2

is:
θ=30∘
Hence, the angle by which the magnet is to be rotated to halve the couple acting on it is:
Option B
30∘